Problem 74
Question
Evaluate the following limits or state that they do not exist. $$\lim _{\theta \rightarrow 0} \frac{\frac{1}{2+\sin \theta}-\frac{1}{2}}{\sin \theta}$$
Step-by-Step Solution
Verified Answer
Question: Evaluate the limit: \(\lim_{\theta\rightarrow0} \frac{\frac{1}{2+\sin\theta}-\frac{1}{2}}{\sin\theta}\)
Answer: The limit of the given expression as \(\theta\) approaches 0 is \(\frac{-1}{4}\).
1Step 1: Simplify the given expression
Simplify the fraction within the limit expression:
$$\frac{\frac{1}{2+\sin\theta}-\frac{1}{2}}{\sin\theta} = \frac{1}{\sin\theta} \cdot \left(\frac{1}{2+\sin\theta}-\frac{1}{2}\right)=\frac{2-(2+\sin\theta)}{\sin\theta(2+\sin\theta)(2)}$$
Now simplify the numerator:
$$\frac{2-(2+\sin\theta)}{\sin\theta(2+\sin\theta)(2)}=\frac{-\sin\theta}{\sin\theta(2+\sin\theta)(2)}$$
2Step 2: Cancel out common factors
Cancel out the common factor \(\sin\theta\) in the numerator and denominator:
$$\frac{-\sin\theta}{\sin\theta(2+\sin\theta)(2)}=\frac{-1}{(2+\sin\theta)(2)}$$
3Step 3: Evaluate the limit
Now, apply the limit:
$$\lim_{\theta\rightarrow0}\frac{-1}{(2+\sin\theta)(2)}=\frac{-1}{(2+\sin(0))(2)}=\frac{-1}{(2+0)(2)}=\frac{-1}{4}$$
So, the limit exists and it is equal to \(\frac{-1}{4}\).
Key Concepts
Trigonometric LimitsL'Hôpital's RuleIndeterminate Forms
Trigonometric Limits
In calculus, trigonometric limits often involve functions like sine, cosine, and tangent. These functions are essential in determining behavior near specific points, like zero. Take, for example, the limit expression involving sine as noted in the exercise. When evaluating \[\lim _{\theta \rightarrow 0} \frac{\frac{1}{2+\sin \theta}-\frac{1}{2}}{\sin \theta}\], the sine function appears in both the numerator and denominator.
To tackle trigonometric limits:
To tackle trigonometric limits:
- First, simplify the expression as much as possible. Look to factor or cancel out common terms.
- Convert the problem into a more familiar form, especially when sine, cosine, or tangent is involved.
L'Hôpital's Rule
L'Hôpital's Rule is a powerful tool in calculus for solving limits where direct substitution leads to an indeterminate form, like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). It involves taking derivatives of the numerator and denominator until an indeterminate form is resolved.
Applying this rule may reveal the true behavior of the fraction as the variable approaches a specific point. To use L'Hôpital's Rule:
Applying this rule may reveal the true behavior of the fraction as the variable approaches a specific point. To use L'Hôpital's Rule:
- First, verify that the limit results in an indeterminate form.
- If valid, differentiate the numerator and the denominator separately.
- Then, re-evaluate the limit with these new expressions.
Indeterminate Forms
When dealing with limits, you’ll often encounter indeterminate forms. These are expressions where direct evaluation doesn't immediately reveal a limit's value. The most common indeterminate forms are \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\).
Consider the initial fraction \[\lim _{\theta \rightarrow 0} \frac{\frac{1}{2+\sin \theta}-\frac{1}{2}}{\sin \theta}\]. Direct substitution of \(\theta = 0\) results in \(\frac{0}{0}\), making it an indeterminate form.
To resolve indeterminate forms:
Consider the initial fraction \[\lim _{\theta \rightarrow 0} \frac{\frac{1}{2+\sin \theta}-\frac{1}{2}}{\sin \theta}\]. Direct substitution of \(\theta = 0\) results in \(\frac{0}{0}\), making it an indeterminate form.
To resolve indeterminate forms:
- Attempt algebraic manipulation, such as factorization or simplification.
- Use tools like L'Hôpital's Rule where differentiation can resolve the form efficiently.
Other exercises in this chapter
Problem 73
Evaluate the following limits or state that they do not exist. $$\lim _{x \rightarrow \pi / 2} \frac{\sin x-1}{\sqrt{\sin x}-1}$$
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Calculate the following limits using the factorization formula \(x^{n}-a^{n}=(x-a)\left(x^{n-1}+x^{n-2} a+x^{n-3} a^{2}+\cdots+x a^{n-2}+a^{n-1}\right)\) where
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Steady states If a function \(f\) represents a system that varies in time, the existence of \(\lim f(t)\) means that the system reaches a steady state (or equil
View solution